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vodomira [7]
3 years ago
9

What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20 to 75

Chemistry
2 answers:
german3 years ago
8 0

Answer:

55°C

Explanation:

original is 20

after gaining heat is 75

75 - 20 = 55

Sindrei [870]3 years ago
7 0

Answer:

17242,5J

Explanation:

Given parameters:

Mass of water  = 75g

Initial temperature  = 20°C

Final temperature  = 75°C

Unknown:

Amount of heat absorbed  = ?

Solution

To find the amount of heat absorbed by water;

      H  = mcΔt

 m is the mass

  c is the specific heat capacity of water  = 4.18J/g°C

 Insert the parameters and solve;

       H  = 75 x 4.18 x (75 - 20)

       H  = 17242,5J

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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

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log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

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mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0
3 years ago
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